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2019 AMC 8 Problems/Problem 9

Revision as of 15:14, 4 January 2023 by Oinava (talk | contribs) (→‎Solution 4)

Problem 9

Alex and Felicia each have cats as pets. Alex buys cat food in cylindrical cans that are $6$ cm in diameter and $12$ cm high. Felicia buys cat food in cylindrical cans that are $12$ cm in diameter and $6$ cm high. What is the ratio of the volume of one of Alex's cans to the volume of one of Felicia's cans?

$\textbf{(A) }1:4\qquad\textbf{(B) }1:2\qquad\textbf{(C) }1:1\qquad\textbf{(D) }2:1\qquad\textbf{(E) }4:1$

Solution 1

Using the formula for the volume of a cylinder, we get Alex, $\pi108$, and Felicia, $\pi216$. We can quickly notice that $\pi$ cancels out on both sides, and that Alex's volume is $1/2$ of Felicia's leaving $1/2 = \boxed{1:2}$ as the answer.

~aopsav

Solution 2

Using the formula for the volume of a cylinder, we get that the volume of Alex's can is $3^2\cdot12\cdot\pi$, and that the volume of Felicia's can is $6^2\cdot6\cdot\pi$. Now, we divide the volume of Alex's can by the volume of Felicia's can, so we get $\frac{1}{2}$, which is $\boxed{\textbf{(B)}\ 1:2}$.

-(Algebruh123)2020

Solution 3

The ratio of the numbers is $1/2$. Looking closely at the formula $r^2 * h * \pi$, we see that the $r * h * \pi$ will cancel, meaning that the ratio of them will be $\frac{1(2)}{2(2)}$ = $\boxed{\textbf{(B)}\ 1:2}$.

-Lcz


Solution 4

The second can is $\cdot 2$ size in each of 2 dimensions, and $\cdot 1/2$ size in 1 dimension. $2^2/2 = \boxed{\textbf{(B)}\ 1:2}$.

~oinava

Video Solution

The Learning Royal : https://youtu.be/8njQzoztDGc

Video Solution by OmegaLearn

https://youtu.be/FDgcLW4frg8?t=2440

~ pi_is_3.14

Video Solution

Solution detailing how to solve the problem: https://www.youtube.com/watch?v=G-gEdWP0S9M&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=10

Video Solution

https://youtu.be/FLT3iOKBC8c

~savannahsolver

See also

2019 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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